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nidhis.1408
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Algebraic Approach:nidhis.1408 wrote:List at least six factors of the product of 3 consecutive even integers.
Say, the first of the 3 even integer is divisible by only 2, i.e. it is of the form 2(2k + 1) = (4k + 2), for some non-negative integer k.
Hence, other two even integers are (4k + 4) and (4k + 6)
Therefore, the product = (4k + 2)(4k + 4)(4k + 6) = [2(2k + 1)]*[4(k + 1)]*[2(2k + 3)] = 16(k + 1)(2k + 1)(2k + 3)
Now, (2k + 1) and (2k + 3) are consecutive odd integers. Hence, one of them must be divisible by 3.
Therefore, 16 and 3 are factors of the product.
Hence, the factors of the product are : 2, 3, 4, 6, 8, 12, 16, 24, and 48
